How fast is $ v_{n+1} = \sqrt[3]{ v_n^3 + v_n^\sqrt 7} ? $ What is a good asymptotic for it ?

I know that $ v_{n+1} = \sqrt[3]{ v_n^3 + 1} $ grows like $ v_{n+1} = \sqrt[3]{ v_1^3 + n} $ and I assume $ v_{n+1} = \sqrt[3]{ v_n^3 + v_n^2} $ grows more or less like $v_1 + c n $ for some fixed $c$ because $ v_{n+1} = \sqrt[3]{ v_n^3 + 3 v_n^2 + 3 v_n + 1} = v_n + 1 $ and those expressions are “close”.

But I started to really wonder about

$$ v_{n+1} = \sqrt[3]{ v_n^3 + v_n^A} ?$$

For various values of $A$.

In particular how fast is

$$ v_{n+1} = \sqrt[3]{ v_n^3 + v_n^\sqrt 7} ? $$

I assume no closed form exists.

Im not sure if using Taylor or Laurent series would help. Or maybe matrices ? The problem can be restated with Taylor , Laurent and matrices. But that does not seem to make it easier at first.

I did some numerical tests but I was not convinced of anything.

Ofcourse if one iterates $v + q(v) $ where $q(v) $ goes to 0 as $v$ increases we could estimate $ n $ iterations to give $ v + \sum_i^n q(i)^{-1} q(v)$.

For example when $q(v) = +/- 1/v $ this works relatively well. But not super good. ( for the negative case take v_1 > 2n )

  • 2
    $\begingroup$ Why not just take $u_n=v_n^3$, when the iteration becomes $u_{n+1}=u_n+u_n^{\sqrt{7}/3}$? I don't see much point to the external cube root... $\endgroup$ Jan 5, 2020 at 22:36

1 Answer 1


I guess that you are assuming that $v_1>0$ (The case $v_1=0$ is trivial, and the $v_n^{\sqrt{7}}$ forces us to work with $v_1\ge 0$).

Now, Notice that $v_n\to\infty$, $v_n$ is increasing, $v_{n+1}/v_n\to 1$ and $$v_{n+1}-v_n = \frac{v_{n+1}^3-v_n^3}{v_n^2+v_n v_{n+1} +v_{n+1}^2} =\frac{v_n^{\sqrt{7}} }{v_n^2+v_n v_{n+1} +v_{n+1}^2}\to \infty $$ From this we can see that $v_n$ grows faster than $cn$.

A slightly well known technique to find the rate of growth of such sequences is mentioned in this small article, we try to find $a$ such that $v_{n+1}^a-v_n^a\to L$ where $L$ is a non zero real number. In this case, The only $a$ that works is $3-\sqrt{7}$ (I have found it using Mathematica, Taylor series can be used to find it). So (By stolz lemma),

$$\frac{1}{n}\sum_{k=1}^n v_{n+1}^a -v_n^a =L= \frac{3-\sqrt{7}}{3}$$ So $v_{n+1}^a\sim L n$. Therefore

$$v_{n} \sim \left(\frac{3-\sqrt{7}}{3} n\right)^{\frac{1}{3-\sqrt{7}}}$$

Here's a plot of $v_n$ in blue and the RHS in orange

enter image description here


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .